Why is taking the dot product with the third vector equivalent to multiplying by the length?

fzeng

How does the determinant formula generalize to arbitrary dimensions?

daria

Where did the dot product come from?

Alan7996

Is this parallelepiped supposed to look this skewed? It seems to be that the diagram suggests that every side looks different, but shouldn't each vertex be either vector u, v, or w?

twizzler

To add to fzeng's question, is there some geometric interpretation of taking the determinant with an arbitrary number of vectors?

Kaxano

For what purpose is it useful to know the volume of the parallelepiped formed by these vectors?

anag

Is there any intuition behind how the algebraic representation of the determinant leads to that geometric interpretation? I get that the determinant satisfies those properties of signed volume, but why is it this way?

frogger

What exactly do we mean by 'signed' volume here? When do we say the volume is positive and when do we say it is negative?

gfkang

How is this meaningful in defining the determinant, since we don't usually use the determinant in this way in its applications?

anon

Why is the sign of the volume important? Also, for the initial u, v, w order in the input of the determinant, how would one choose which order is the right one for their calculations?

spidey

Though we may not actually use it in practice, how would we find the determinant for dimensions greater than 3, and what would that determinant mean in terms of the matrix for these larger dimensions?

bunnybun99

Why does computing the determinant of three vectors equal to a cross product and a dot product operation?

Coyote

Is volume ever going to be negative? I would think probably not for physical simulations and the like, right? Also can the triple product formula ever generalize out or do we have to switch back to calculating sub-determinants after the third dimension?

Why is taking the dot product with the third vector equivalent to multiplying by the length?

How does the determinant formula generalize to arbitrary dimensions?

Where did the dot product come from?

Is this parallelepiped supposed to look this skewed? It seems to be that the diagram suggests that every side looks different, but shouldn't each vertex be either vector u, v, or w?

To add to fzeng's question, is there some geometric interpretation of taking the determinant with an arbitrary number of vectors?

For what purpose is it useful to know the volume of the parallelepiped formed by these vectors?

Is there any intuition behind how the algebraic representation of the determinant leads to that geometric interpretation? I get that the determinant satisfies those properties of signed volume, but why is it this way?

What exactly do we mean by 'signed' volume here? When do we say the volume is positive and when do we say it is negative?

How is this meaningful in defining the determinant, since we don't usually use the determinant in this way in its applications?

Why is the sign of the volume important? Also, for the initial u, v, w order in the input of the determinant, how would one choose which order is the right one for their calculations?

Though we may not actually use it in practice, how would we find the determinant for dimensions greater than 3, and what would that determinant mean in terms of the matrix for these larger dimensions?

Why does computing the determinant of three vectors equal to a cross product and a dot product operation?

Is volume ever going to be negative? I would think probably not for physical simulations and the like, right? Also can the triple product formula ever generalize out or do we have to switch back to calculating sub-determinants after the third dimension?

What is a parallelpiped?

Why do we use dot product here?

what is the physical meaning of triple product?